Problems On Numbers

Let the ten's digit be x
Then, unit's digit = x + 3
Number = 10x + (x + 3)
              = 11x + 3
Sum of digit's = x + (x + 3)
                      = 2x + 3
$$\therefore \frac{{11x + 3}}{{2x + 3}} = \frac{4}{1}$$
$$ ⇒ 11x + 3 = 8x + 12$$
$$ ⇒ 3x = 9$$
$$ ⇒ x = 3$$
Hence, Required number
= 11x + 3
= 11 × 3 + 3
= 36

Let the numbers be×and (x + 3)
Then,
$$ ⇒ {x^2} + {\left( {x + 3} \right)^2} = 369$$
$$ ⇒ {x^2} + {x^2} + 9 + 6x = 369$$
$$ ⇒ 2{x^2} + 6x - 360 = 0$$
$$ ⇒ {x^2} + 3x - 180 = 0$$
$$ ⇒ \left( {x + 15} \right)\left( {x - 12} \right) = 0$$
$$ ⇒ x = 12$$
So, the numbers are 12 and 15
∴ Required sum = (12 + 15) = 27

Details

Let the ten's digit be x and unit's digit be y
Then, x + y = 15 and x - y = 3   or   y - x = 3
Solving x + y = 15   and   x - y = 3, we get: x = 9, y = 6
Solving x + y = 15   and   y - x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69
Hence, the number cannot be determined.

Let the numbers be a, b and c
Then, a² + b² + c² = 138 and (ab + bc + ca) = 131
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) = 138 + 2×131 = 400
⇒ (a + b + c) = $$\sqrt {400} $$  = 20

Details

Details

Details

Let the ten's digit be x.

Then, unit's digit = x + 2.

Number = 10x + (x + 2) = 11x + 2

Sum of digit's = x + (x + 2) = 2x + 2.

(11x + 2) (2x + 2) = 144

⇒ 22x^2+26x-140=0

⇒ (x - 2)(11x + 35) = 0

⇒ x = 2

Hence, Required Number = 11x + 2 = 24

Let the required number be x
Then,
$$ ⇒ x + 20 = \frac{{69}}{x}$$
$$ ⇒ {x^2} + 20x - 69 = 0$$
$$ ⇒ {x^2} + 23x - 3x - 69 = 0$$
$$ ⇒ x\left( {x + 23} \right) - 3\left( {x + 23} \right) = 0$$
$$ ⇒ \left( {x + 23} \right)\left( {x - 3} \right) = 0$$
$$ ⇒ x = 3\,\,\,\,\,\,\,\,\,\left[ {\because×\ne - 23} \right]$$